Closed Topologist's Sine Curve is Connected

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Let $G$ be the graph of the function $y = \map \sin {\dfrac 1 x}$ for $x > 0$.

Let $J$ be the line segment joining the points $\tuple {0, -1}$ and $\tuple {0, 1}$ in $\R^2$.

Then $G \cup J$ is connected.


Since the open interval $\openint 0 \infty$ is connected, then so is $G$ by Continuous Image of Connected Space is Connected.

It is enough, from Set between Connected Set and Closure is Connected, to show that $J \subseteq \map \cl G$.

Let $p \in J$, say, $\tuple {0, y}$ where $-1 \le y \le 1$.

We need to show that:

$\forall \epsilon > 0: \map {N_\epsilon} p \cap G \ne \O$

where $\map {N_\epsilon} p$ is the $\epsilon$-neighborhood of $p$.

Let us choose $n \in \N: \dfrac 1 {2 n \pi} < \epsilon$.

From Sine of Half-Integer Multiple of Pi:

$\map \sin {\dfrac {\paren {4 n + 1} \pi} 2} = 1$


$\map \sin {\dfrac {\paren {4 n + 3} \pi} 2} = -1$

So by the Intermediate Value Theorem, $\map \sin {\dfrac 1 x}$ takes every value between $-1$ and $1$ in the closed interval $\closedint {\dfrac 2 {\paren {4 n + 3} \pi} } {\dfrac 2 {\paren {4 n + 1} \pi} }$.

In particular, $\map \sin {\dfrac 1 {x_0} } = y$ for some $x_0$ in this interval.

The distance between the points $\tuple {0, y}$ and $\tuple {x_0, \map \sin {\dfrac 1 {x_0} } } = \tuple {x_0, y}$ is $x_0 < \epsilon$.


$\tuple {x_0, \map \sin {\dfrac 1 {x_0} } } \in \map {N_\epsilon} p \cap G$

as required.